Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $a \neq 0$. $q = \dfrac{-a - 9}{4a^2 + 8a} \div \dfrac{2a^2 + 6a - 108}{a^3 - 3a^2 - 18a} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{-a - 9}{4a^2 + 8a} \times \dfrac{a^3 - 3a^2 - 18a}{2a^2 + 6a - 108} $ First factor out any common factors. $q = \dfrac{-(a + 9)}{4a(a + 2)} \times \dfrac{a(a^2 - 3a - 18)}{2(a^2 + 3a - 54)} $ Then factor the quadratic expressions. $q = \dfrac {-(a + 9)} {4a(a + 2)} \times \dfrac {a(a - 6)(a + 3)} {2(a - 6)(a + 9)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac {-(a + 9) \times a(a - 6)(a + 3) } {4a(a + 2) \times 2(a - 6)(a + 9) } $ $q = \dfrac {-a(a - 6)(a + 3)(a + 9)} {8a(a - 6)(a + 9)(a + 2)} $ Notice that $(a - 6)$ and $(a + 9)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {-a\cancel{(a - 6)}(a + 3)(a + 9)} {8a\cancel{(a - 6)}(a + 9)(a + 2)} $ We are dividing by $a - 6$ , so $a - 6 \neq 0$ Therefore, $a \neq 6$ $q = \dfrac {-a\cancel{(a - 6)}(a + 3)\cancel{(a + 9)}} {8a\cancel{(a - 6)}\cancel{(a + 9)}(a + 2)} $ We are dividing by $a + 9$ , so $a + 9 \neq 0$ Therefore, $a \neq -9$ $q = \dfrac {-a(a + 3)} {8a(a + 2)} $ $ q = \dfrac{-(a + 3)}{8(a + 2)}; a \neq 6; a \neq -9 $